Bh The normal boiling point of water is 100.0 degreeC and its molar enthalpy of
ID: 915994 • Letter: B
Question
Bh The normal boiling point of water is 100.0 degreeC and its molar enthalpy of vaporization is 40.67 kJ/mol. What is the change in entropy in the system in J/K when 24.7 grams of steam at 1 atm condenses to a liquid at the normal boiling point? -150 -40.7 -88.8 373 88.8 Of the following, the entropy of gaseous is the largest at 25 degreeC and 1 atm. H2 C3H4 C2H6 C3H8 C3H6 The standard Gibbs free energy of formation of is zero. H_2O(l) Fe(s) I2(s) (a) only (b) only Q (c) only (b) and (c) (a), (b), and (c)Explanation / Answer
20) The enthalpy of vap. = 40.67 kJ/mol = 40.67 kJ/ 18 g of water [as mol. wt. of water is 18]
Therefore for condensation of 24.7 g of water the enthalpy change = - (40.67 x 24.7/ 18) kJ = - 55.81 kJ
therefore entropy change = enthalpy change / temperature = - 55.81 kJ/ 373 K = -149.6 J/K
So (A) is the answer
21) (D) is the answer. As entropies generally increase with molecular weight
22) (D) is the answer. Standard Gibbs free energy of formation zero for elements in their elemental state.
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